Decimal expansion of rational numbers

Correct answer: A. (37)

Explanation: (0.\overline{045}=\frac{45}{999}=\frac{5}{111}), so the denominator is (111). An initial zero inside the repeating block is also counted as a digit.

Correct answer: A. (\frac{5}{111})

Explanation: (0.\overline{045}=\frac{45}{999}), and reducing by (9) gives (\frac{5}{111}). First form the denominator with (9)'s according to the repeating digits.

Correct answer: A. Terminating after (4) places

Explanation: Both (3^0) and (17^0) equal (1), so the effective denominator is (2^4\cdot 5^3). The larger exponent is (4), so the decimal terminates after (4) places.

Correct answer: C. (6)

Explanation: The factor (31) makes the decimal recurring, and the larger exponent of (2) and (5) is (6), giving the non-repeating start. In mixed denominators, the larger exponent gives the delay.

Correct answer: A. (2^4\cdot 5)

Explanation: (0.01875=\frac{1875}{100000}=\frac{3}{160}), and (160=2^5\cdot 5). The correct prime factorisation is (2^5\cdot 5), so complete the calculation before choosing.

Correct answer: A. (\frac{3}{160})

Explanation: (0.01875=\frac{1875}{100000}), and dividing by (625) gives (\frac{3}{160}). Convert the decimal to a fraction and reduce fully.

Correct answer: B. Non-terminating recurring

Explanation: A rational number has a terminating decimal only when, after cancellation, its denominator has no prime factors other than 2 and 5. If another prime factor remains, the division cannot end; because remainders repeat, the decimal becomes non-terminating recurring. This rule helps classify the decimal without carrying out a long division.

Here, the numerator is 200 = \(2^3\times5^2\). Cancelling common factors with \(2^3\times5^3\times7\) leaves the denominator \(5\times7\), since one factor 5 and the factor 7 remain. The factor 7 is not allowed in a terminating denominator, so the decimal is non-terminating recurring. Therefore option B follows.

Correct answer: A. (\frac{58}{2^3\cdot 5^2\cdot 29})

Explanation: Since (58=2\cdot 29), the factor (29) cancels and the reduced denominator is (2^2\cdot 5^2). If an extra prime appears, check cancellation first.

Correct answer: A. Terminating

Explanation: (0.\overline{27}=\frac{27}{99}) and (0.\overline{72}=\frac{72}{99}), so their sum is (1). The sum of two recurring decimals can be terminating.

Correct answer: B. Non-terminating recurring with (3) initial non-repeating digits

Explanation: Since (11) remains, the decimal is non-terminating recurring. The larger exponent in (2^3\cdot 5^3) gives (3) initial non-repeating digits.

Correct answer: A. Only (2) and (5) can occur

Explanation: For a terminating decimal, the reduced denominator (q) can contain only (2) and (5). In (q^3), powers increase but no new prime factor appears.

Correct answer: B. Non-terminating recurring

Explanation: A rational number has a terminating decimal only when, after reducing the fraction to lowest terms, its denominator has no prime factors other than 2 and 5. If any other prime remains in the denominator, the decimal division cannot end; because the remainders eventually repeat, the decimal is non-terminating recurring.

Here, cancel the common factors in the numerator and denominator: \(2^4\) cancels part of \(2^7\), and one factor 13 cancels part of \(13^2\). The reduced denominator is \(2^3\cdot 5^3\cdot 13\). Since the prime factor 13 remains, the decimal expansion is non-terminating recurring. Therefore, option B is correct; it is not terminating and cannot be non-recurring because the number is rational.

Correct answer: B. (32)

Explanation: (0.03125=\frac{3125}{100000}=\frac{1}{32}). Convert a terminating decimal to a fraction and reduce the denominator.

Correct answer: A. (\frac{1}{3200})

Explanation: (0.0003125=\frac{3125}{10000000}), and reducing by (3125) gives (\frac{1}{3200}). Do not forget to cancel common factors in large denominators.

Correct answer: C. (\frac{1}{224})

Explanation: (224=2^5\cdot 7), so (5) non-repeating digits appear before the recurring part. For comparison, check the larger power of (2) and (5).

Correct answer: B. 175

Explanation: Reason: \(10^6=2^6\cdot 5^6\). The given denominator has \(2^6\) but only \(5^4\), so it is short by \(5^2\). Multiply numerator and denominator by \(5^2=25\) to get denominator \(10^6\). Thus \(N=7\times25=175\). Note on distractors: 35 comes from multiplying by \(5^1\) (a common slip), 700 from multiplying by 100, and 875 from multiplying by \(5^3=125\). Exam tip: compare prime-power factors of the denominator with those of \(10^n\) to find the exact factor to multiply quickly.

Correct answer: B. Non-terminating recurring

Explanation: After cancellation, the denominator is (2^3\cdot 3\cdot 5^4\cdot 11), which contains (3) and (11). If primes other than (2) and (5) remain in the reduced denominator, the decimal is non-terminating recurring.

Correct answer: C. (2375)

Explanation: Since (10^5=2^5\cdot 5^5), the denominator lacks (5^3). Therefore (N=19\cdot 125=2375); multiply by the missing factor when making a power of (10).

Correct answer: B. (1998)

Explanation: One non-repeating zero and three repeating digits give (\frac{125}{9990}), which reduces to (\frac{25}{1998}). In mixed recurring decimals, do not treat the first denominator as the final one.

Correct answer: B. Terminates exactly after (6) places

Explanation: Since (q) divides (10^6), it has only (2) and (5), but not dividing (10^5) means the larger exponent is (6). Therefore the decimal terminates exactly after (6) places.

Correct answer: C. (9) places

Explanation: The denominator has only (2) and (5), so the decimal terminates with the larger exponent (9). In exams, use the larger exponent instead of adding exponents.

Correct answer: B. (3) places

Explanation: Since (484=2^2\cdot 11^2), the reduced denominator is (2^2\cdot 5^3). The larger exponent is (3), so reduce first and then count decimal places.

Correct answer: C. (2601)

Explanation: For termination, (3^2) and (17^2) must cancel completely, so (n=2601). For the least value, cancel only the factors other than (2) and (5).

Correct answer: B. (110)

Explanation: (0.4272727\ldots=\frac{423}{990}=\frac{47}{110}), so the denominator is (110). In mixed recurring decimals, the final fraction must be reduced.

Correct answer: B. (37)

Explanation: (0.\overline{108}=\frac{108}{999}=\frac{4}{37}). First form the denominator with (9)'s according to the repeating digits and then reduce.